Subgroup of hyperbolic group generated by non-torsion elements Let $G$ be a hyperbolic group. I know that it is an open problem whether $G$ has a torsion-free subgroup of finite index. But if we let $N$ be the subgroup of $G$ generated by its non-torsion elements, then is $G/N$ necessarily finite?
If not, $G/N$ would be a finitely generated infinite torsion group with finitely many conjugacy classes.
 A: Yes. The following answer is inspired by Andy Putman's comment. Let $N_\infty(G)$ be the subgroup generated by elements of infinite order, in a group $G$.
Every non-elementary hyperbolic group $G$ with trivial finite radical has a useful property, which bears the ridiculous name "Pnaive", namely that for every finite subset $F$ there exists a $x$ such that the natural homomorphism from $F\ast\mathbf{Z}$ to $G$ mapping $F$ identically to itself and the generator of $\mathbf{Z}$ to $x$, is injective. This is a result of Arzhantseva and Minasyan: G. Arzhantseva, A. Minasyan, Relatively hyperbolic groups are C*-simple. 
J. Funct. Anal. 243 (2007), no. 1, 345–351.
This applies to $F$ equal to an arbitrary singleton $\{y\}$, $y\in G$: then both $x$ and $yx$ have infinite order, hence $y\in N_\infty(G)$. So Pnaive for a group $G$ implies $G=N_\infty(G)$, that is, $G/N_\infty(G)=\{1\}$.
Moreover, the property $N_\infty(G)=G$ for nontrivial (!) groups, is also stable under taking extensions with torsion kernel. Hence it is satisfied by all non-elementary hyperbolic groups, since these have a maximal finite normal subgroup and we inherit Pnaive from the nontrivial quotient.
For completeness, the elementary cases: if $G$ is finite then $N_\infty(G)=\{1\}$ and the quotient $G/N_\infty(G)$ is $G$ itself. If $G$ is 2-ended, then, denoting by $W(G)$ the finite radical, $G/W(G)$ is infinite cyclic or infinite dihedral according to whether $G$ acts trivially or not on its 2-element boundary, and then the desired quotient $G/N_\infty(G)$ is either trivial, or cyclic of order 2. 
Thus for an arbitrary hyperbolic group, $G/N_\infty(G)$ is finite.
(Note that we use much much less than the full result of Arzhantseva-Minasyan! we only need it for a singleton $\{y\}$, and we need much less than injectivity: only injectivity on the cyclic subgroup $\langle xy\rangle$.)
A: Here is how I would prove it using ping-pong. The relevant form of ping-pong is the following proposition:
Proposition. Let $X$ be a set and $G_1, G_2$ be groups acting on $X$ such that there exist two subsets $X_1, X_2\subset X$ with disjoint complements and nonempty intersection, such that 
$$
\forall g_i\in G_i -\{1\}, g_i(X_i)\cap X_i=\emptyset. 
$$
Then the group $\langle G_1, G_2\rangle$ of bijections of $X$ generated by $G_1, G_2$ is naturally isomorphic to $G_1\star G_2$. 
Now, back to our hyperbolic group $G$. I will assume that $G$ is a non elementary hyperbolic group whose finite radical $W(G)$ is trivial. (Otherwise, see Yves' answer.) Given $g\in G$, let $Fix(g)$ denote the fixed-point set of $g$ in $\partial G$ (the Gromov boundary of $G$). 
The following lemma answers Yves' question: 
Lemma 1. If $g\in G-\{1\}$ has finite order then $Fix(g)$ has empty interior. 
Proof. Suppose not. Then, since $W(G)=1$, $Fix(g)\ne \partial G$. 
Recall:
Theorem. The set of pairs $(p_+, p_-)$ of attractive/repulsive fixed points of infinite order elements of $G$ is dense in $\partial G\times \partial G$. 
Hence, one can find such fixed pair for an element $h\in G$ such that $p_+\notin Fix(g), p_-\in Fix(g)$. Then the sequence of conjugates
$$
g_n=h^n g h^{-n}, n\in {\mathbb N},
$$ 
contains infinitely many distinct elements (since the sequence of complements to $Fix(g_n)$ in $\partial G$ converges to the singleton $p_+$).  At the same time, the sequence $(h^{-n})_{n\in  {\mathbb N}}$ is uniformly close to the quasiconvex hull of $Fix(g)$. Since $g$ acts on this quasiconvex hull with bounded displacement, we obtain that the sequence of word norms $|g_n|$ is bounded. A contradiction. qed
Lemma 2. Let $g\in G$ be nontrivial,  generating a finite cyclic group $C< G$. Let $h\in G$ be an infinite order element such that 
$$
g^k(Fix(h))\cap Fix(h)=\emptyset
$$
for all $g^k\in C-\{1\}$. Then there exists $n$ such that the subgroup $\langle g, h^n\rangle$ of $G$ generated by $g$ and $h^n$ is naturally isomorphic to $C \star \langle h^n \rangle$. 
Proof. Let $U$ be a small neighborhood of $Fix(h)$ in $\partial G$ such that 
$$
g^k(U)\cap U=\emptyset
$$
for all $g^k\in C-\{1\}$ (such a neighborhood existents since $C$ is finite). Let $V= \partial G -U$. Then there exists $n\in {\mathbb N}$ such that for all nontrivial elements $f\in \langle h^n\rangle$ we have
$$
f(V)\cap V=\emptyset. 
$$
It now follows from Tits' ping-pong that the subgroup $\langle g, h^n\rangle$ is naturally isomorphic to $C \star \langle h^n\rangle $. qed 
Lemma 3. Given a nontrivial finite order element $g\in G$, there exists an infinite order element $h\in G$ such that $$
g^k(Fix(h))\cap Fix(h)=\emptyset
$$
for all $g^k\in \langle g \rangle  -\{1\}$.
Proof. Let $U$ be a nonempty open subset of $\partial G -U$. Since $C=\langle g\rangle $ is finite, we can choose $U$ such that 
for all nontrivial elements $f\in C$ we have
$$
f(U)\cap U=\emptyset. 
$$
(For instance, one can use Lemma 1 here.) 
On the other hand, by  minimality of the action of $G$ on $\partial G$, 
there exists an infinite order element  $h\in G$ such that $Fix(h)\subset U$. (This also follows from the Theorem stated above.) qed 
Now, Lemmas 2 and 3 imply:
Corollary. Let $g\in G$ be a nontrivial finite order element. 
Then there exists an infinite order element $h\in G$ such that the subgroup $\langle g, h\rangle$ of $G$ is naturally isomorphic to $\langle g\rangle \star \langle h\rangle$. In particular, the product $gh$ also has infinite order. 
